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%%文档的题目、作者与日期
\author{王立庆（2019级数学与应用数学1班）}
\title{数量金融实验 - 第5章课堂练习}
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%\date{2022 年 9 月 8 日}

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\item %1
下述哪个不是蒙特卡洛方法的创始人之一？
\begin{enumerate}
\item[A.]  Stanislaw Marcin Ulam
\item[B.]  Enrico Fermi
\item[C.]  John von Neumann
\item[D.]  Nicholas Metropolis
\item[E.]  Monte Carlo
\end{enumerate}

{\color{red}
解答：E. Monte Carlo is a city in Monaco. 

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\item %2
设 $X_1,\cdots,X_n$ 是来自均匀分布 $U[0,1]$ 的简单随机样本。设 $\overline{X}$ 是样本均值。则 $\overline{X}$ 的方差是多少？
\begin{enumerate}
\item[A.]  $1/n$
\item[B.]  $1/(2n)$
\item[C.]  $1/(4n)$
\item[D.]  $1/4$
\end{enumerate}

{\color{red}
解答：C. 计算均值和方差，可知，
\begin{eqnarray*}
\mathbb{E}(X_i) &=& \frac{1}{2}, \\ 
\mathbb{E}(X_i^2) &=& \frac{1}{2}, \\ 
Var(X_i) &=& \mathbb{E}(X_i^2) - \mathbb{E}(X_i)^2 = \frac{1}{4}, \\
Var(\overline{X}) &=& \frac{1}{n^2} Var(X_1+\cdots+X_n) =\frac{1}{n^2} [Var(X_1)+\cdots+Var(X_n)] = \frac{1}{4n}.
\end{eqnarray*}

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\item %3
设有二维随机变量 $(X,Y)$ 的一个随机样本 $(x_i,y_i)$. 其协方差的估计表达式是哪个？
\begin{enumerate}
\item[A.]  $\sum\limits_{i=1}^{n} x_i \sum\limits_{i=1}^{n} y_i$ 
\item[B.]  $\sum\limits_{i=1}^{n} x_iy_i$. 
\item[C.]  $\sum\limits_{i=1}^{n} (x_i-\bar{x})(y_i-\bar{y})$
\item[D.]  $\frac{1}{n-1}\sum\limits_{i=1}^{n} (x_i-\bar{x})(y_i-\bar{y})$
\end{enumerate}

{\color{red}
解答：D. 协方差 $cov(X,Y)$ 的估计式是 $$\frac{1}{n-1}\sum\limits_{i=1}^{n} (x_i-\bar{x})(y_i-\bar{y}).$$

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\item %4
设有随机变量 $X$ 与函数 $f(x)$. 记 $A=\mathbb{E}[f(X)]$, 它的蒙特卡洛估计表达式为 
$$\hat{A}_N = \frac{1}{N}\sum\limits_{i=1}^{N} f(X_i).$$
为了提高估计的精度，可以下述说法中，不正确的是哪个？
\begin{enumerate}
\item[A.]  可以使用控制变量方法来减小这个估计表达式的方差。
\item[B.]  可以使用重要抽样方法来减小这个估计表达式的方差。
\item[C.]  可以使用最小二乘蒙特卡洛方法来减小这个估计表达式的方差。
\item[D.]  可以使用增加 $N$ 来减小这个估计表达式的方差。
\end{enumerate}

{\color{red}
解答：C. 最小二乘蒙特卡洛方法结合线性回归和蒙特卡洛方法，没提到估计精度这方面的考虑。

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\item %5
关于有限差分方法，下述说法中，不正确的是哪个？
\begin{enumerate}
\item[A.]  显式差分格式总是收敛的。
\item[B.]  隐式差分格式需要求解包含很多未知数的线性方程组。
\item[C.]  Crank-Nicolson差分格式也称为六点差分格式。
\item[D.]  有限差分方法的基本思想是用差分代替微分。
\end{enumerate}

{\color{red}
解答：A. 显式差分格式的收敛是有条件的。

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